Question

Prove that if x, and y are vectors in a unitary space, then
4(x,y) = ||x+y||^2-||x-y||^2+i||x+iy||^2-i||x-iy||^2

Answer 1

unitary space?

Answer 2

yes, it said so

Answer 3

define it.

Answer 4

That's just inner product in Complex space. The definition on the book says: The real inner product space is sometimes called a Euclidean space; its complex analogue is called a unitary space.

Answer 5

and defined by (x,y ) = \(\sum_{i=1}^n x_i\bar y_i\)

Answer 6

Okay, sue that definition then

Answer 7

use^

Answer 8

Maybe try to simplify the right side...

Answer 9

yes, I did and get so ugly result. :)

Answer 10

\[
\|x+y\|^2 = (x+y, x+y)
\]

Answer 11

the first term ||x+y||^2 = ((x+y), (x+y) ) =(x,x) + (x,y)+(y,x) + (y,y)

Answer 12

the second term ||x-y||^2 = (x,x) - (x,y) -(y,x)+(y,y)

Answer 13

am I correct so far?

Answer 14

Ummm

Answer 15

Is it true that (a+b,c) = (a,c) + (b,c)?

Answer 16

yes, the second axiom said that

Answer 17

oh, no, it says so for inner product, not for any vectors

Answer 18

\[
(a+b,c) = \overline{(c,a+b)} = \overline{(c,a) + (c,b)} = \overline{(c,a)} + \overline{(c,b)} = (a,c) +(b,c)
\]You use this?

Answer 19

I thought second axiom just says (a,b+c) = (a,b) + (a,c).

Answer 20

Ummm, please restate axiom.

Answer 21

ok, so you can do it for (x+y,x+y) = (x,x+y) + (y,x+y).

Answer 22

But can you do the other half?

Answer 23

At the bottom, they fix the first term, and expand exactly what we did. right?

Answer 24

just don't have alpha. To the problem it is 1(real), right?

Answer 25

ok

Answer 26

Well then... continue it.

Answer 27

the 3rd term
i||x + iy||^2 = i[(x+iy),(x+iy)]= i[(x,x) +(x,iy)+(iy,x)+(iy,iy)]
=i(x,x)+i(x,iy)+i(iy,x) +i(iy,iy)

Answer 28

the last one:
i||x-iy||^2= i[(x-iy),(x-iy) ]= i[(x,x)-(x,iy)-(iy,x)+(iy, iy)]
=i(x,x)-i(x,iy)-i(iy,x)+i(iy,iy)

Answer 29

copy all
(x,x) + (x,y)+(y,x) + (y,y) -(x,x) + (x,y) +(y,x)-(y,y)+i(x,x)+i(x,iy)+i(iy,x) +i(iy,iy)-i(x,x)+i(x,iy)+i(iy,x)-i(iy,iy)

Answer 30

simplify
2(x,y)+2(y,x)+2i(x,iy)+2i(iy,x)

Answer 31

and stuck, hehehe... don't know how to do next

Answer 32

maybe axiom 2 can be used on the is.

Answer 33

2(y,x)+2i(iy,x) = (2y + 2iiy, x)

Answer 34

is it not ii = -1?

Answer 35

ok

Answer 36

Can I do: 2(y,x)+2i(iy,x) = (2y + 2iiy, x) =0
so that the left is 2(x,y) +2i(x, iy)
for the second one, apply the unitary space axiom on iy, take i out, it becomes \(\bar i\), so that I have 2(x,y) + 2i \(\bar i\)(x,y) = 4(x,y)